Question

On a given condition , the equilibrium concentration of $HI$, $H_2$ and $I_2$ are $0.80,0.10and0.10mole/litre$ the equilibrium constant for the reaction $H_2$+$I_2$$\rightleftharpoons 2HI$ will be:

• A. $64$
• B. $12$
• C. $8$
• D. $0.8$

Answer 1

The correct option is A $64$

The reaction given,
$H_2+I_2\rightleftharpoons 2HI;\left[HI\right]=0.80,\left[H_2\right]=0.10,\left[I_2\right]=0.10$
$\Rightarrow K_c=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}\Rightarrow \frac{0.80\times 0.80}{0.10\times 0.10}=64$